# Does a sequence $a_n$ converge if $|a_n-\frac{1}{n} \sum_{i=1}^n a_i| \to 0$, as $n \to \infty$?

Let $$\{a_n\}$$ be a real sequence. If $$\lim\limits_{n\to \infty} \left|a_n - \frac{1}{n}\sum\limits_{i=1}^n a_i \right|= 0$$, do we have $$\lim\limits_{n\to \infty} a_n$$ convergent?

This is somehow an inverse version of Cesaro mean convergence, which says that if $$\lim\limits_{n\to \infty} a_n = L$$, then $$\lim\limits_{n\to \infty} \frac{1}{n}\sum\limits_{i=1}^n a_i = L$$. See

On cesaro convergence: If $x_n \to x$ then $z_n = \frac{x_1 + \dots +x_n}{n} \to x$

I first tried $$a_n = \ln(n)$$. $$\frac{1}{n}\sum_{i=1}^n a_i =\ln(n!)/n \approx (n\ln(n)-n+O(\ln(n)))/n =\ln(n)-1+o(1)$$. Close.

Looking at something slower growing, I tried $$a_n = \ln\ln(n+1)$$ and this works.

$$\int \ln(\ln(x))dx =x\ln(\ln(x))+\int \dfrac{dx}{\ln(x)} =x\ln(\ln(x))+\dfrac{x}{\ln(x)}+O(\dfrac{x}{\ln^2(x)})$$

By Euler-Maclaurin,

$$\begin{array}\\ \sum_{k=2}^n \ln\ln(k) &=\int_2^n \ln\ln(x)dx+\dfrac{\ln\ln(n)+\ln\ln(2)}{2}+O((\ln\ln(n))')\\ &=n\ln\ln(n)+\dfrac{n}{\ln(n)}+O(\dfrac{n}{\ln^2(n)})+\dfrac{\ln\ln(n)+\ln\ln(2)}{2}+O(\dfrac1{n\ln(n)})\\ \text{so}\\ \dfrac{\sum_{k=2}^n \ln\ln(k)}{n} &=\ln\ln(n)+\dfrac{1}{\ln(n)}+O(\dfrac{1}{\ln^2(n)})+\dfrac{\ln\ln(n)+\ln\ln(2)}{n}+O(\dfrac1{n^2\ln(n)})\\ &=\ln\ln(n)+o(1)\\ \end{array}$$

Therefore the limit is zero and $$a_n$$ diverges.

• What if the sequence is bounded, i.e., there exists $M$ such that $\|a_n\| \leq M, \; \forall n \geq 1$ ? – Allen Zhang Apr 18 at 7:50
• Yes there is. See my second answer. – marty cohen Apr 18 at 21:01

No: consider for example $$a_n=\log\log n$$.

Edited to add: go upvote marty cohen's answer for working out the details!

• Easy so say, herder to prove. – marty cohen Apr 17 at 21:02
• Thanks for doing the work! – Greg Martin Apr 17 at 22:02
• Also @martycohen: I still feel like this is a somewhat incomplete solution: can it happen that $a_n$ converges neither to a finite limit, nor to infinity? – W-t-P Apr 18 at 5:51
• I think an example like $\sin(\log\log x)$ would work for that. – Greg Martin Apr 18 at 6:08
• @W-t-P: I have added a second answer showing that such a $a_n$ exists (i.e., bounded and does not converge). – marty cohen Apr 18 at 20:03

This is a second answer because I am turning the problem completely around.

I will show that there is a sequence $$a_n$$ such that $$\lim_{n\to \infty} \big|a_n - \frac{1}{n}\sum_{i=1}^n a_i \big|= 0$$, $$a_n$$ is bounded, and $$a_n$$ does not converge.

$$\lim_{n\to \infty} \big|a_n - \frac{1}{n}\sum_{i=1}^n a_i \big|= 0$$ means that $$a_n =\frac{1}{n}\sum_{i=1}^n a_i +f(n)$$ where $$f(n) \to 0$$ as $$n \to \infty$$.

Then $$na_n =\sum_{i=1}^n a_i +nf(n)$$ or $$(n-1)a_n =\sum_{i=1}^{n-1} a_i +nf(n)$$.

Applying the usual trick when $$\sum_{i=1}^n a_i$$ appears in a recurrence, $$na_{n+1} =\sum_{i=1}^{n} a_i +(n+1)f(n+1)$$.

Subtracting,

$$\begin{array}\\ na_{n+1}-(n-1)a_n &=\sum_{i=1}^{n} a_i +(n+1)f(n+1) -(\sum_{i=1}^{n-1} a_i +nf(n))\\ &=a_n +(n+1)f(n+1) -nf(n)\\ \end{array}$$

so

$$\begin{array}\\ na_{n+1} &=na_n +(n+1)f(n+1) -nf(n)\\ \text{or}\\ a_{n+1} &=a_n +(1+1/n)f(n+1) -f(n)\\ \end{array}$$

Therefore $$a_{n+1}-a_n =(1+1/n)f(n+1) -f(n) =f(n+1) -f(n)+\frac{f(n+1)}{n}$$.

Summing,

$$\begin{array}\\ a_n-a_1 &=\sum_{k=1}^{n-1} (a_{k+1}-a_k)\\ &=\sum_{k=1}^{n-1} (f(k+1) -f(k)+\dfrac{f(k+1)}{k})\\ &=f(n)-f(1)+\sum_{k=2}^{n} \dfrac{f(k)}{k-1}\\ \end{array}$$

Now, by choosing a $$f(k) \to 0$$, we can get a desired $$a_n$$.

However, if we want $$a_n \to \infty$$, we must also have $$\sum_{k=2}^{n} \dfrac{f(k)}{k-1} \to \infty$$.

$$f(k) = \frac1{k}$$ will not work because the sum converges.

$$f(k) = \frac1{\ln(k)}$$ works because $$\sum_{k=2}^n \dfrac{1}{\ln(k)(k-1)} \approx \sum_{k=2}^n \dfrac{1}{k\ln(k)} \approx \int_{k=2}^n \dfrac{dx}{x\ln(x)} = \ln(\ln(n))$$.

If we want an $$a_n$$ that doesn't converge and stays bounded, we must find a $$f(k)$$ such that $$\sum_{k=2}^{n} \dfrac{f(k)}{k-1}$$ behaves like that.

Since $$\sum_{k=2}^n \dfrac{1}{\ln(k)(k-1)} \approx \ln(\ln(n))$$, for any $$n_0$$ there is a $$n(n_0)$$ such that $$\sum_{k=n_0}^{n(n_0)} \dfrac{1}{\ln(k)(k-1)} \gt 1$$. An approximate value is $$n(n_0) \approx n_0^e$$.

Therefore, if we choose $$f(k)$$ to have constant sign in each interval $$n_0 \le k \lt n(n_0)$$, with the signs alternating in consecutive intervals, as $$n$$ increases $$\sum_{k=2}^{n} \dfrac{(-1)^{g(k)}}{\ln(k)(k-1)}$$ for the appropriate $$g(k)$$ will increase by 1, then decrease by 1, and so on, and will thus not converge and be bounded.